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Rotational Dynamics

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Did you know that people, no matter where in the world, all experience rotational dynamics? As a result of our planet’s constant rotation about its axis, we are all moving at about \( 1,600\,\mathrm{\frac{km}{h}}! \) Earth rotates about its vertical axis, which is tilted 23.5 degrees, and completes one full rotation after 24 hours. Moreover, since the seasons are the result of different parts of the earth facing in the direction of the sun, it is because of this rotation that we experience different seasons. Therefore, with the importance of rotational dynamics being undeniable, let's introduce some definitions and examples that will help us expand our knowledge on the topic.

Let's start by clarifying what we refer to as rotational dynamics.

**Rotational dynamics **is the** **study of rotational motion resulting from external forces and their torques.

This area of mechanics is very broad, but we will try to give it a general overview.

We can describe rotational motion using four main physical quantities: angular velocity, angular acceleration, angular displacement, and time. While we have an intuitive notion of time, the other rotational motion variables might not sound that familiar to you, so let's start by defining each of them.

**Angular displacement,** \( \Delta \theta \),** **is the difference between an initial and final angular position around a specified axis.

The mathematical formula corresponding to this definition is

$$\Delta \theta= \theta_f - \theta_i.$$

Angular displacement has an SI unit of radians (\(\mathrm{rad}\)).

**Angular velocity, \( \omega \),** is the rate of an object's change in angular displacement with respect to time.

The mathematical formula for average angular velocity is

$$\omega=\frac{\Delta{\theta}}{\Delta{t}},$$

where \( \Delta\theta \) is angular displacement and \( t \) is time. Angular velocity has a SI unit of \( \mathrm{\frac{rad}{s}} \). However, for instantaneous angular velocity we use

$$\omega=\frac{\mathrm{d}\theta}{\mathrm{d}t}.$$

**Angular acceleration, \( \alpha \),** is an object's change in angular velocity with respect to time.

We can calculate the average angular acceleration as

$$\alpha=\frac{\Delta{\omega}}{\Delta{t}},$$

and for instantaneous angular acceleration

$$ \alpha=\frac{\mathrm{d}\omega}{\mathrm{ d}t} .$$

Angular acceleration has a unit of \( \mathrm{\frac{rad}{s^2}} \).

Now besides understanding these definitions and corresponding formulas, one should also be aware of the relationship between these quantities. Angular velocity is the first derivative of angular displacement with respect to time, \( \omega=\frac{d\theta}{dt} \), while angular acceleration is the first derivative of angular velocity with respect to time, \( \alpha=\frac{d\omega}{dt} \) as well as the second derivative of angular displacement \( \alpha=\frac{d^2\theta}{dt^2}. \)

Given the angular position function, \( \theta(t)=\left(5\;\mathrm{\frac{rad}{s^2}}\right) t^2 +\left(8\;\mathrm{\frac{rad}{s^2}}\right)t -6\;\mathrm{rad}, \) calculate the angular velocity and angular acceleration functions.

**Solution **

To find the angular velocity function we can calculate the derivate of the angular position function with respect to time.

\begin{align}\omega(t)&=\frac{\mathrm{d}\theta}{\mathrm{d}t}\\\omega(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left( \left(5\;\mathrm{\frac{rad}{s^2}}\right) t^2 +\left(8\;\mathrm{\frac{rad}{s}}\right)t -6\;\mathrm{rad}\right)\\\omega(t)&=\left(10\;\mathrm{\frac{rad}{s^2}}\right)t+8\;\mathrm{\frac{rad}{s}}\end{align}

Similarly, since angular acceleration is the second derivative of the angular position function, we can find it by calculating the derivative of the angular velocity function.

\begin{align}\\\alpha(t)&=\frac{\mathrm{d}\omega}{\mathrm{d}t}\\\alpha(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left(\left(10\;\mathrm{\frac{rad}{s^2}}\right)t+8\;\mathrm{\frac{rad}{s}}\right)\\\alpha(t)&=10\;\mathrm{\frac{rad}{s^2}}\\\end{align}

A radian is an angle measure frequently used in a circular motion.

A **radian** is the angle corresponding to an arc length equal to the circle's radius.

If we want to convert radians to degrees, we can use the formula,

$$\theta_\text{degrees}=\theta_\text{radians}\left(\frac{180^\circ}{\pi\;\mathrm{rad}} \right). $$

On the other hand, to convert from degrees to radians, we can apply the formula:

$$\theta_\text{radians}=\theta_\text{degrees}\left(\frac{\pi\,\mathrm{rad}}{180^\circ}\right).$$ Let's check our understanding with a quick example.

Convert the following quantities:

- 5.00 radians into degrees
- 275 degrees into radians

**Solution**

We use the corresponding formula according to the units of the given angle.

\begin{align}\theta_\text{degrees}&=\theta_\text{radians}\left(\frac{180^\circ}{\pi\;\mathrm{rad}} \right)\\\theta_\text{degrees}&=5.00\,\text{rad}\left(\frac{180^\circ}{\pi\,\mathrm{rad}}\right)\\\theta_\text{degrees}&=\frac{(5.00)180^\circ}{3.14}\\\theta_\text{degrees}&=286^\circ\\\end{align}

\begin{align}\theta_\text{radians}&=\theta_\text{degrees}\left(\frac{\pi\,\mathrm{rad}}{180^\circ}\right)\\\theta_\text{radians}&=275^\circ\left(\frac{\pi\,\mathrm{rad}}{180^\circ}\right)\\\theta_\text{radians}&=\frac{(275^\circ)(\pi\;\mathrm{rad})}{180^\circ}\\\theta_\text{radians}&=4.80\,\text{radians}\\\end{align}

Note that \( 1\,\text{radian} \approx 57.3^\circ.\)

Torque is the rotational equivalent of a force. For an object to experience angular acceleration, it has to be affected by torque.

**Torque**, \( \tau \), is a vector quantity that quantifies the turning effect of a force applied to an object.

The SI unit for torque is \( \mathrm{N\,m.} \)

Convention states that a counterclockwise rotation indicates positive torque, and a clockwise rotation indicates negative torque. The amount of torque applied to an object depends on the applied force but also on the perpendicular distance from where the force is applied, with respect to the axis of rotation.

For a system to be at rotational equilibrium, the sum of all torques acting on a system must equal zero.

$$\sum \tau=0$$

The sum of all torques acting on a system can be zero if the torques are acting in opposite directions, thus canceling out.

**Rotational equilibrium** is defined as a state in which neither a system's state of motion nor its internal energy state changes with respect to time.

And we also have a quantity that plays the role of the mass in translational motion. For rotational motion, we use the moment of inertia or rotational inertia.

**Rotational inertia **is a quantitative measurement of an object's resistance to angular acceleration.

Rotational inertia varies depending on the shape of the object and its mass distribution with respect to the axis of rotation.

Now let's talk about the equations of the concepts used in rotational motion and and how we can use them.

Sometimes, it is enough to work with rotational kinematic equations when studying rotational motion.

**Rotational kinematics** refers to the study of rotational motion without considering external forces.

We can use the following kinematic equations whenever we are dealing with a constant angular acceleration.

Angular velocity equation:

$$\omega=\omega_{o} + \alpha{t}.$$

Angular displacement equation:

$$\Delta{\theta} =\omega_o{t}+\frac{1}{2}{\alpha}t^2.$$

Angular velocity squared equation:

$$\omega^2={\omega_{o}}^2 +2{\alpha}\Delta{\theta}.$$

In the above equations, \(\omega\) is final angular velocity, \(\omega_0\) is the initial angular velocity, \(\alpha\) is angular acceleration, \(t\) is time, and \(\Delta \theta\) is angular displacement.

Now, as we have discussed rotational dynamics and its connection to rotational kinematics, we must be sure to also recognize and understand the relationship between rotational kinematics and linear kinematics. These relationships appear in the table below, where \( r \) is the radius of the circular motion.

Variable | Linear Abbreviation | Linear SI units | Angular Abbreviation | Angular SI units | Relationship |

acceleration | $$a$$ | $$\frac{m}{s^2}$$ | $$\alpha$$ | $$\frac{rad}{s^2}$$ | $$a=\alpha{r}, \alpha=\frac{a}{r}$$ |

velocity | $$v$$ | $$\frac{m}{s}$$ | $$\omega$$ | $$\frac{rad}{s}$$ | $$v=\omega{r}, \omega=\frac{v}{r}$$ |

displacement | $$x$$ | $$m$$ | $$\Delta \theta$$ | $$rad$$ | $$x=\theta{r}, \theta=\frac{x}{r}$$ |

time | $$t$$ | $$s$$ | $$t$$ | $$s$$ | $$t=t$$ |

To better understand these relationships, let us look at the diagram below.

Torque can be calculated by using three different formulas.

- Cross Product Formula
- Magnitude Formula
- Newtons Second Law Formula

The cross-product definition of torque is expressed by the equation

$$\tau=\vec{r} \times \vec{F}$$

where \( \vec{r} \) is the displacement from the axis of rotation and \( \vec{F} \) is the applied force.

The cross product is also known as the vector product. Its result is another vector quantity whose direction is perpendicular to both vector arguments and the plane defined by them. The torque vector is parallel to the axis of rotation.

If we only care about the magnitude of the torque, \( | \vec{\tau} | = \tau, \) we can calculate it using the following formula:

$$ \tau = | \vec r \times \vec F | = r F \sin\theta.$$

where \( r \) is the magnitude of the displacement from the axis of rotation, \( F \) is the magnitude of the applied force, and \( \theta \) is the angle between them.

We can also write the above equation as

$$ \tau = Fr_\perp,$$

where \( r_\perp = r\sin\theta ,\) is the perpendicular distance from the axis of rotation to the line of action of the force, known as the lever arm.

The equation for torque can be written in the same format as that of Newton's second law, \( F=ma \).

$$\tau=I\alpha$$

Here \( I \) is the moment of inertia and \( \alpha \) is the magnitude of the angular acceleration.

The moment of inertia for a point mass at a distance of \(r\) from the axis of rotation is

$$I=mr^2.$$

This formula is important because we can use it to derive formulas for the moment of inertia of complex systems since objects can be thought of as being comprised of sets of point masses. The total rotational inertia for a collection of objects is calculated by the formula

$$I_{\mathrm{total}}= \sum I_i=\sum m_ir_{i}^2.$$

Let's try a quick example.

A point mass of \( 2.1\,\mathrm{kg} \) is positioned \( 0.89\,\mathrm{m} \) from the axis of rotation. Calculate its moment of inertia.

**Solution**

We are given both the mass and the radius. Therefore, we can apply the moment of inertia formula for a point mass directly.

$$\begin{align}I&=mr^2\\I&=(2.1\,\mathrm{kg})(0.89\,\mathrm{m})^2\\I& = 1.7\,\mathrm{kg\,m^2}\\\end{align}$$

If our system is a finite-sized body, we can model it as being comprised of infinitely many differential masses \( \mathrm{d}m \). Each of these masses contributes to the moment of inertia of the object, but instead of adding them up, we need to integrate them to find the total moment of inertia of the object:

$$I=\int{r^2}dm,$$

where \( r \) is the perpendicular distance from the differential masses, \( dm,\) to the axis of rotation.

We still have many other counterparts to those of translational motion. Let's talk briefly about a few of them.

Just as velocity is used to calculate kinetic energy,

$$K=\frac{1}{2}mv^2,$$

we can quantify the rotational kinetic energy of a system knowing its angular velocity

$$K_{\text{rot}}= \frac{1}{2}I\omega^2.$$

**Angular momentum** is a vector quantity defined as the product of angular velocity and rotational inertia.

The mathematical formula corresponding to this definition is

$$L=I\omega,$$

where \( \omega \) is angular velocity measured in \( \mathrm{\frac{rad}{s}}, \) and \( I \) is inertia measured in \( \mathrm{kg\,m^2}. \) Angular momentum has SI units of \( \mathrm{kg\,\frac{m^2}{s}} \).

This formula can only be used when the moment of inertia is constant.

We also have a definition for the angular impulse which describes how torque, the rotational analog of force, affects a system with respect to time.

**Angular impulse** is defined as the product of the torque exerted on a system over a specific time interval.

The mathematical formula corresponding to this definition is

$$\Delta \vec{J}_{rot}= \int_{t_o}^{t}\vec{\tau}(t)dt,$$

which can be simplified to

$$J=\tau\Delta{t},$$

when \( \tau \) doesn't vary with time. Since \( \tau \) is measured in \( \mathrm{N\,m} \) and \( t \) in \( \mathrm{s}, \) the angular impulse unit in the SI system is \( \mathrm{N\, \frac{m}{s}}. \)

Angular impulse and angular momentum are related by the impulse-momentum theorem:

$$ J=\Delta{L} .$$

To demonstrate the concept of rotational dynamics in the lab, one can complete an experiment involving a rotating arm. In this experiment, the apparatus consists of a rotating arm connected to a rotary motion sensor that measures angular velocity. On the rotating arm, two masses will be attached to slow it down by increasing its rotational inertia. The further out the masses are, the harder it is for the arm to rotate. The axis of the rotating arm will then be attached to a string which is also attached to a pulley with a hanging mass. This setup allows for a constant torque to be applied to the arm, because of tension from the string, causing it to rotate. Torque can then be measured and continuously varied using different hanging masses.

To solve rotational dynamic problems, one can use a combination of the equations we explored before. Let us apply our new-found knowledge of rotational dynamics to the following two examples.

The spinning wheel has an initial angular velocity of \( 5.0\,\mathrm{\frac{rad}{s}}.\) If its angular acceleration is given as \(2.1\,\mathrm{\frac{rad}{s^2}} \) and its angular displacement is given as \( 10.6\,\mathrm{rad},\) what is the final angular velocity of the Ferris wheel? How long does it take to reach this velocity?

**Solution**

Note that we are given the following information:

- initial angular velocity
- angular acceleration
- angular displacement

Hence, we can identify and use the equation,

$$\omega^2={\omega_o}^2+ 2\alpha\Delta\theta,$$

to solve the first part of this problem. Therefore, our calculations are as follows:

$$\begin{aligned} \omega^2 &= \omega_0^2 + 2\alpha\Delta\theta\\ \omega^2&=\left( 5.0\,\mathrm{\frac{rad}{s}} \right)^2+ 2\left( 2.1 \mathrm{\frac{rad}{s^2}} \right) \left( 10.6\,\mathrm{rad}\right)\\ \omega^2&=25 + 44.52\\ \omega^2&=69.52\\ \omega&=8.3\,\mathrm{\frac{rad}{s}} \end{aligned}$$

Now using this information and the equation,\( \Delta\theta=\frac{\omega + \omega_o}{2}{t}, \) we can calculate the time it takes to reach this final velocity value as follows:

$$\begin{align}\Delta\theta&=\frac{\omega + \omega_o}{2}{t}\\10.6\,\mathrm{rad}&=\frac{8.3\,\mathrm{\frac{rad}{s} + 5.0\frac{rad}{s}}}{2}(t)\\10.6&= 6.7(\mathrm{t})\\t&= \frac{10.6}{6.7}=1.6\,\mathrm{s}\\\end{align}$$

The spinning wheel reaches its final angular velocity of \( 8.3\,\mathrm{\frac{rad}{s}} \) at \( t=1.6\,\mathrm{s}. \)

Now let's try another example.

A \( 5\,\mathrm{kg} \) soccer ball, initially at rest, is spun and after \( 2.4\,\mathrm{s} \) is rotating with an angular velocity of \( 4.5\,\mathrm{\frac{rad}{s}}. \) Calculate the ball's angular acceleration. If the radius is given to be \( 0.7\,\mathrm{m}, \) calculate the torque exerted on the ball. Note that a soccer ball is considered a thin spherical shell with a corresponding rotational inertia formula of \( I=\frac{2}{3}mr^2. \)

**Solution**

Based on the problem, we are given

- initial and final angular velocity
- time

Therefore, we can apply our equation, \(\omega = \omega_0 + \alpha t\), to solve the first part of this problem. First, we must rearrange our equation to solve for angular acceleration. \begin{align*} \omega &= \omega_0 + \alpha t \\ \alpha t &= \omega - \omega_0 \\ \alpha &= \frac{\omega - \omega_0}{t} \end{align*} We can now insert our variables and solve as follows: \begin{align*} \alpha &= \frac{\omega - \omega_0}{t} \\ &= \frac{4.5\,\frac{\mathrm{rad}}{\mathrm{s}}-0\,\frac{\mathrm{rad}}{\mathrm{s}}}{2.4\,\mathrm{s}} \\ &= 1.9\,\frac{\mathrm{rad}}{\mathrm{s}^2} \end{align*} For the second part of this problem, we must apply the equation,\( \tau=I{\alpha}.\) First we must rewrite the torque equation using the equation given for the moment of inertia, $$\begin{align}\tau&=\left(\frac{2}{3}mr^2\right)\alpha\\\end{align}$$ and now we can insert our values. $$\begin{align}\tau&=\left(\frac{2}{3}mr^2\right)\alpha\\\tau&=\left(\frac{2}{3}(5\,\mathrm{kg})(0.7\,\mathrm{m})^2\right)1.9\,\frac{\mathrm{rad}}{\mathrm{s}^2}\\\tau&=3.1\,\mathrm{N\,m}\\\end{align}$$

- Rotational dynamics is the study of motion and the forces causing objects to rotate about an axis.
- Rotational motion
- Rotational kinematics
- There are three rotational kinematic equations:
- The angular velocity equation \( \omega=\omega_{o} + \alpha{t}, \)
- The angular displacement equation \( \Delta{\theta} =\omega_o{t}+\frac{1}{2}{\alpha}t^2, \)
- The angular velocity squared equation \( \omega^2={\omega_{o}}^2 +2{\alpha}\Delta{\theta}.\)

- Torque is a vector quantity, that quantifies the turning effect of a force applied to an object.
- We can calculate torque using three different formulas:
- \( \tau=\vec{r} \times \vec{F} \)
- \( \tau=rF\sin\theta\)
- \( \tau=I\alpha\)

- The moment of inertia is the measurement of an object's resistance to angular acceleration.
- An object's inertial formula can be expressed as \( I=mr^2 \) for a point mass and \( I=\int{r^2}dm \) for a finite body-sized system.
- Rotational equilibrium
- Rotational equilibrium is achieved when the sum of all toques in a system is zero.
- Angular momentum is a vector quantity defined as the product of angular velocity and rotational inertia. Its corresponding formula is \( L=I\omega. \)
- Angular impulse is defined as the product of the torque exerted on a system over a specific time interval.

- Fig. 1 - Radian diagram, StudySmarter Originals
- Fig. 2 - Relationship diagram for kinematic variables, StudySmarter Originals
- Fig. 3 - Torque diagram, StudySmarter Originals
- Fig. 4 - Rotational Dynamics Apparatus, StudySmarter Originals

** **study of rotational motion resulting from external forces and their torques.

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